RelationBetweenForceAndPotentialEnergy 2006-07-14 10:32:00 2009-03-06 18:14:35 Definition % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % define commands here \usepackage{amsmath, amssymb, amsfonts, amsthm, amscd, latexsym} \usepackage{xypic} \usepackage[mathscr]{eucal} \theoremstyle{plain} \newtheorem{lemma}{Lemma}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}{Corollary}[section] \theoremstyle{definition} \newtheorem{definition}{Definition}[section] \newtheorem{example}{Example}[section] %\theoremstyle{remark} \newtheorem{remark}{Remark}[section] \newtheorem*{notation}{Notation} \newtheorem*{claim}{Claim} \renewcommand{\thefootnote}{\ensuremath{\fnsymbol{footnote%%@ }}} \numberwithin{equation}{section} \newcommand{\Ad}{{\rm Ad}} \newcommand{\Aut}{{\rm Aut}} \newcommand{\Cl}{{\rm Cl}} \newcommand{\Co}{{\rm Co}} \newcommand{\DES}{{\rm DES}} \newcommand{\Diff}{{\rm Diff}} \newcommand{\Dom}{{\rm Dom}} \newcommand{\Hol}{{\rm Hol}} \newcommand{\Mon}{{\rm Mon}} \newcommand{\Hom}{{\rm Hom}} \newcommand{\Ker}{{\rm Ker}} \newcommand{\Ind}{{\rm Ind}} \newcommand{\IM}{{\rm Im}} \newcommand{\Is}{{\rm Is}} \newcommand{\ID}{{\rm id}} \newcommand{\GL}{{\rm GL}} \newcommand{\Iso}{{\rm Iso}} \newcommand{\Sem}{{\rm Sem}} \newcommand{\St}{{\rm St}} \newcommand{\Sym}{{\rm Sym}} \newcommand{\SU}{{\rm SU}} \newcommand{\Tor}{{\rm Tor}} \newcommand{\U}{{\rm U}} \newcommand{\A}{\mathcal A} \newcommand{\Ce}{\mathcal C} \newcommand{\D}{\mathcal D} \newcommand{\E}{\mathcal E} \newcommand{\F}{\mathcal F} \newcommand{\G}{\mathcal G} \newcommand{\Q}{\mathcal Q} \newcommand{\R}{\mathcal R} \newcommand{\cS}{\mathcal S} \newcommand{\cU}{\mathcal U} \newcommand{\W}{\mathcal W} \newcommand{\bA}{\mathbb{A}} \newcommand{\bB}{\mathbb{B}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bD}{\mathbb{D}} \newcommand{\bE}{\mathbb{E}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bG}{\mathbb{G}} \newcommand{\bK}{\mathbb{K}} \newcommand{\bM}{\mathbb{M}} \newcommand{\bN}{\mathbb{N}} \newcommand{\bO}{\mathbb{O}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bV}{\mathbb{V}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bfE}{\mathbf{E}} \newcommand{\bfX}{\mathbf{X}} \newcommand{\bfY}{\mathbf{Y}} \newcommand{\bfZ}{\mathbf{Z}} \renewcommand{\O}{\Omega} \renewcommand{\o}{\omega} \newcommand{\vp}{\varphi} \newcommand{\vep}{\varepsilon} \newcommand{\diag}{{\rm diag}} \newcommand{\grp}{{\mathbb G}} \newcommand{\dgrp}{{\mathbb D}} \newcommand{\desp}{{\mathbb D^{\rm{es}}}} \newcommand{\Geod}{{\rm Geod}} \newcommand{\geod}{{\rm geod}} \newcommand{\hgr}{{\mathbb H}} \newcommand{\mgr}{{\mathbb M}} \newcommand{\ob}{{\rm Ob}} \newcommand{\obg}{{\rm Ob(\mathbb G)}} \newcommand{\obgp}{{\rm Ob(\mathbb G')}} \newcommand{\obh}{{\rm Ob(\mathbb H)}} \newcommand{\Osmooth}{{\Omega^{\infty}(X,*)}} \newcommand{\ghomotop}{{\rho_2^{\square}}} \newcommand{\gcalp}{{\mathbb G(\mathcal P)}} \newcommand{\rf}{{R_{\mathcal F}}} \newcommand{\glob}{{\rm glob}} \newcommand{\loc}{{\rm loc}} \newcommand{\TOP}{{\rm TOP}} \newcommand{\wti}{\widetilde} \newcommand{\what}{\widehat} \renewcommand{\a}{\alpha} \newcommand{\be}{\beta} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta} \newcommand{\del}{\partial} \newcommand{\ka}{\kappa} \newcommand{\si}{\sigma} \newcommand{\ta}{\tau} \newcommand{\lra}{{\longrightarrow}} \newcommand{\ra}{{\rightarrow}} \newcommand{\rat}{{\rightarrowtail}} \newcommand{\oset}[1]{\overset {#1}{\ra}} \newcommand{\osetl}[1]{\overset {#1}{\lra}} \newcommand{\hr}{{\hookrightarrow}} \textit{This is a contributed entry} \section{Potential Energy and Force acting on a Particle} Let us asssume from the start that the field's force $\mathbf{F}$ is irrotational, i.e. $\nabla \times \mathbf{F}= \mathbf{0}$, that is, $\nabla \times \mathbf{F}=\mathbf{0} \leftrightarrow \mathbf{F}=-\nabla U$. In another words, the field's force is conservative if and only if it is irrotational. So the conseravation of mechanical energy $dE/dt=d(T+U)/dt=0$ is a consequence of that theorem. Once one imposes $\nabla \times \mathbf{F}=\mathbf{0}$, then one is proving that the necessary condition for $\mathbf{F}=-\nabla U$. Another consequence about the theorem is that the ``work'' of the field's force is independent of the path described by the particle in its motion. That is, if $\Gamma_1$ and $\Gamma_2$ are two different paths, described by the particle, and joininig its initial and end position on the time interval $[t_1,t_2]$, then the line integrals $\int_{\Gamma_1}\mathbf{F}\cdot d\mathbf{r}= \int_{\Gamma_2}\mathbf{F}\cdot d\mathbf{r}$ must be equal and hence the work of the field's force, as the particle describes a closed path, must be zero, i.e. $\oint\mathbf{F}\cdot d\mathbf{r}=0$. The relation between the force, $\mathbf{F}$, acting on a particle, and the potential energy, $U$ of that particle is: \begin{equation} \mathbf{F} = -\nabla U, \end{equation} where $\nabla$ is the gradient operator. \subsection{Derivation} The above relationship can be derived from the conservation of energy. Let $T$ denote the kinetic energy of a particle, and $U$ its potential energy, with $E$ the total energy, given by $E=T+U$. Take the total time derivative of $E$, giving \begin{equation} \frac{dE}{dt} = \frac{dT}{dt} + \frac{dU}{dT} \end{equation} The kinetic energy of a particle is expressed as $T=\frac{1}{2}mv^{2}$, where $m$ is the mass of the particle, and $v$ is the magnitude of the particle's velocity. Recall that by Newton's second law, $\mathbf{F} = md\mathbf{v}/dt$, where $\mathbf{v}$ is the velocity vector. Consider, next, the quantity $\mathbf{F}\cdot d\mathbf{r}$, where $\mathbf{r}$ is the position vector of the particle. Expanding $\mathbf{F}$ in terms of Newton's second law, it is seen that \begin{eqnarray*} \mathbf{F}\cdot d\mathbf{r} & = & m\frac{d\mathbf{v}}{dt}\cdot\frac{d\mathbf{r}}{dt}dt\\ & = & m\frac{d\mathbf{v}}{dt}\cdot\mathbf{v}dt\\ & = & \frac{1}{2}m\frac{d}{dt}(\mathbf{v}\cdot\mathbf{v})dt\\ & = & d(\frac{1}{2}mv^{2}) = dT. \end{eqnarray*} Therefore, $dT/dt = \mathbf{F}\cdot d\mathbf{r}/dt$. It is assumed that the potential energy is a function of time and space \emph{i.e.} $U=U(x_{1},x_{2},x_{3},t)$. The time derivative of the potential energy can be expanded through the chain rule as \begin{equation} \frac{dU}{dt} = \frac{\partial U}{\partial t} + \frac{\partial U}{\partial x_{1}}\frac{\partial x_{1}}{\partial t} + \frac{\partial U}{\partial x_{2}}\frac{\partial x_{2}}{\partial t} + \frac{\partial U}{\partial x_{3}}\frac{\partial x_{3}}{\partial t}. \end{equation} Notice that \begin{equation} \frac{\partial U}{\partial x_{1}}\frac{\partial x_{1}}{\partial t} + \frac{\partial U}{\partial x_{2}}\frac{\partial x_{2}}{\partial t} + \frac{\partial U}{\partial x_{3}}\frac{\partial x_{3}}{\partial t} = \nabla U\cdot\mathbf{v}, \end{equation} and substitute this result, as well as the expression for the time derivative of kinetic energy back into the original equation for the time derivative of the total energy, \begin{eqnarray} \frac{dE}{dt} & = & \frac{dT}{dt} + \frac{dU}{dt}\\ & = & \mathbf{F}\cdot\frac{d\mathbf{r}}{dt} + \frac{\partial U}{\partial t} + \nabla U\cdot\mathbf{v}\\ & = & (\mathbf{F} + \nabla U)\cdot\mathbf{v} + \frac{\partial U}{\partial t} \end{eqnarray} If the potential has no explicity time dependence \emph{i.e.} it is dependent upon position, which is dependent on time, then $dU/dt=0$, and the above becomes \begin{equation} \frac{dE}{dt} = (\mathbf{F} + \nabla U)\cdot\mathbf{v} = 0, \end{equation} where $dE/dt=0$ arises because of the conservation of energy within a closed system \emph{i.e.} energy does not enter or leave the system. Therefore, it follows that under the conservation of energy, and the time independence of potential energy, $\mathbf{F} + \nabla U = 0$, which can be rewritten as \begin{equation} \mathbf{F} = -\nabla U, \end{equation} which is the desired relation between the force acting on a particle, and its potential energy.